Hello, sebasto!

The domain gave me headaches, so I ignored it at the beginning.

Equate the partial derivative to zero and solve.Find the maximum and minimum for in a given domain.

. .

. .

From [2], we have: .

If , [1] becomes: .

. . We have two critical points: .

If , [1] becomes: .

. . We have one more critical point: .

Second Partial Test: .

At . . . saddle points

At . . . extremum

. . . . . positive, concave up

Therefore: . . . . which is in the domain.

Edit: We must find the maximum(s) along the circumference of that circle.

I would use Lagrange multipliers: .